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Usimov [2.4K]
3 years ago
13

Swagelok Enterprises is a manufacturer of miniature fittings and valves. Over a 5-year period, the costs associated with one pro

duct line were as follows: first cost of $20,000, and annual costs of $16,000. Annual revenue was $29,000 and the used equipment was salvaged for $10,000. What rate of return did the company make on this product?
Business
1 answer:
Mars2501 [29]3 years ago
8 0

Answer:

The rate of return did the company make on this product is 61.93%

Explanation:

initial cost = 20000

Annual benefit = 29000

Annual cost = 16000

Salvage value = 10000

t = 5yrs

Let rate of return is I, then at rate of return NPV=0

Present value = -20000 + (29000 - 16000)*(P/A, i, 5) + 10000 (P/F, i, 5)

-20000 + 13000*(P/A, i, 5) + 10000 (P/F, i, 5) = 0

13*(P/A, i, 5) + 10* (P/F, i, 5) = 20

We need to use trail and error method to find rate of return

At = 10%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 55.49

At = 15%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 48.55

At = 25%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 38.24

At = 35%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 31.09

At = 45%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 25.94

At = 55%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 22.11

At = 60%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 20.55

At = 62%, value of expression 13*(P/A, i, 5) + 10* (P/F, i, 5) is 19.98

using interpolation

rate of return = 0.6 + (20.55-20)/(20.55-19.98) * (0.62-0.6)

                       = 0.6+0.019298

                       = 0.6193

                       = 61.93%

Therefore, The rate of return did the company make on this product is 61.93%

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Allen Construction purchased a crane 6 years ago for $130,000. They need a crane of this capacity for the next 5 years. Normal o
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<u>For retaining of Old Machine Equipment</u>

Price of old equipment 3 yrs ago = $130,000

O & M cost per year = $35,000

Using the Cash flow approach

End of year   Cash flow 1   Old equipment

0                            $0            Initial Cash flow

1                         -$35,000     O & M cost per year

2                        -$35,000     O & M cost per year

3                        -$35,000     O & M cost per year

4                        -$35,000     O & M cost per year

5                        -$35,000     O & M cost per year

Hence, Annual worth = Initial cash flow + Annual cost

Annual worth = 0 - $35,000

Annual worth = -$35,000

<u>For buying of new equipment</u>

Cost of buying new crane = $150,000

Market value of old crane = $40,000

Time = 5 years

O & M cost per year = $8,000

Salvage value = $55,000

MARR = 20%

Using the Cash flow approach

End of year   Cash flow 1   New equipment

0                         $110,000    -$150,000 + $40,000

1                         -$8,000     O & M cost per year

2                        -$8,000     O & M cost per year

3                        -$8,000     O & M cost per year

4                        -$8,000     O & M cost per year

5                        $47,000     -$8,000 + $55,000

Annual worth = Initial cash flow + Annual cost + Salvage value

Annual worth = -$110,000(A/P 20%,5) - $8,000 + $55,000(A/P 20%,5)

Annual worth = -$110,000*(0.334) - $8,000 + $55,000*(0.134)

Annual worth = -$36,781.77 - $8,000 + $7,390.88

Annual worth = -$37,908.88

Conclusion: We should retain the old machine as it is more favorable than purchase of new equipment

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